Question

Question 13 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Class X1 - Maths -Sequences and Series Page 192

Answer 1

$Here\:GP\:series\:are:3,3^2,3^3,3^4,.......$

a = 3 , r = 3 and Sn = 120
Now,
$S_n=\frac{a(r^n-1)}{r-1},r\geq1\\\\120=\frac{3(3^n-1)}{3-1}\\\\120=\frac{3(3^n-1)}{2}\\\\120\times2=3(3^n-1)\\240=3(3^n-1)\\80=3^n-1\\81=3^n\\(3)^4=(3)^n\\n=4$

Answer 2

ANSWER:

n=4

EXPLAIN

using formula of SN when R is is greater than 1.

a(r^n-1)/r-1